Hold'em Odds

[jtr]

Hello, Laldaas.

Good question. The 10.5:1 figure that gets thrown around for the gutshot straight, and equivalent odds figures for other situations, represent your chance of hitting the hand you want. So if your call would put you all-in, the logic would be easy: you need odds of 10.5:1 to break even, in which case a call would be EV-neutral. (Folding would also be fine.) Anything better than 10.5:1 would make it +EV to call. This is assuming your draw was to the nuts.

However, real life is rarely that simple. Often your draw is not to the nuts, in which case you can make it and still get beaten, so you'd want significantly better odds than 10.5:1. Sometimes much better if there's a good chance of splitting the pot even when you win, for example. However, the reason most people ask themselves "Am I getting 10.5:1?" and then call if so, is that they figure to make at least a bet or two on the end when their draw hits. Whereas they will fold when it doesn't come in. So even though calling a gutshot draw at exactly 10.5:1 looks like an EV-neutral move, it's really a positive move as the implied odds are better than the pot odds (which just means there's a good chance of more money going in when you hit).

Certainly I think you're right to suspect that people sometimes get lazy about this, and do silly things like calling a gutshot draw at about 9:1 even though their opponent is all-in. But that's just bad poker.

[Ric]

To extend this question for flop play, are EV and pot/implied odds so close that it doesn't matter which you use for hold 'em? You don't hit as often on the turn, but you get paid off about double. Conversely, if you call the turn unimproved but hit the river, you get paid off some there too. It seems that these two different conditions would help even each other out to around where EV is in the first place. Am I right in that assumption?